# Falling body geodesics seem counterintuitive

We all know, a geodesic is the path an object follows to maximize its proper time. Geodesic equations of an object near a gravitational field show that it should fall towards the ground, which is the expected behaviour. However, this is counterintuitive. Clocks tick slower near the ground, so an apple will age slower if it falls than if it floats or even soars upwards.

I know that the geodesic an apple chooses maximizes its proper time between the tree and the ground. But it could have chosen other directions, which would maximize its proper time even more. What causes it to take this particular geodesic? Please notice that, if I am standing on earth I have countless geodesics to choose from at every direction.

Now you can say that the equations dictate that, the free fall geodesics say that if you have no initial spatial velocity, only a temporal velocity, you must fall towards the ground. My question is, why? Why not some other direction where you can age even more?

• Because falling straight done just is the only geodesic passing through your location in space-time that has purely time-like 4-velocity? Jul 26, 2021 at 12:34
• And even if that is, why should it point towards the ground, even though there are 'better' places to go to maximize the proper age? Jul 26, 2021 at 12:38
• "Clocks tick slower...", that doesn't really have anything to do with proper time. Jul 26, 2021 at 17:08

Geodesic equations of an object near a gravitational field shows that it should fall towards the ground, which is the expected behaviour. However, this is counterintuitive. Clocks tick slower near the ground, so an apple will age slower if it falls than if it floats or even soars upwards.

Actually, there are upwards moving geodesics as well as downwards moving geodesics. I believe that the confusion you are having is seen in your "a geodesic is the path an object follows to maximize it’s proper time". This should actually be "a geodesic is the path an object follows between two specified events to maximize it’s proper time".

The laws of physics, as you may be aware, are expressed as second-order differential equations. When you solve a physics problem, you need to not only solve the differential equations, but you must also provide the initial/boundary conditions. Since the equations are 2nd order, you generally need to supply two conditions for each generalized coordinate. These can be either a beginning position and a beginning velocity or a beginning position and an ending position. The "maximize its proper time" formulation of a geodesic is based on the approach using a beginning position and an ending position.

So, consider two events in a uniform gravitational field where the two events are both at the same location, but a given fixed time apart. One path between those two events could be for the object to simply hover in place, but as you said clocks run slower near the ground so you could get more proper time by going upwards a bit and then coming back down. So you might think that if going upwards a bit is good then going upwards a lot would be even better, but as you go upwards a lot your speed increases and you start to get kinematic time dilation slowing your clock. It turns out that there is a maximum, a path which goes as high as possible without going very fast so as to maximize the proper time. That path is (approximately) the traditional parabolic path. Similarly with other pairs of events.

Now, we have seen that in a uniform gravitational field a geodesic path between two events at the same height, a given time apart, must start with an upward velocity, and that the resulting geodesic forms a standard parabola. There is a theorem that shows that a maximal path between two points is also maximal for any other pair of points on the path. This can be used to find other geodesic paths between different pairs of events, and for converting between the initial point + final point and the initial point + initial velocity formulations. Since every parabola has pairs of points at the same level, we can use that to find that the parabola is still the geodesic between pairs of points not at the same level. And since each point on a parabola has a given velocity, we can see that the parabola also defines the geodesic for given initial events with initial velocities.

For example, if we take the apex of the parabola we find it is an event where the object is momentarily at rest (four-velocity purely in the time direction). If we examine all different pairs of initial/final events we find that the apex is always the only event where the object is momentarily at rest. So if we specify any initial event and the initial velocity is at rest, then we know that the initial event is the apex of a parabola. Since it is the apex, then all other points on the parabola must be lower. Therefore, all geodesics starting from rest will go downward.

Please notice that, if I am standing on earth I have countless geodesics to choose from at every direction.

This is correct. This again goes back to the initial/boundary conditions. In simple cases picking an initial point and a final point (for maximizing proper time) is equivalent to picking an initial point and an initial velocity. Every different choice of a final position is equivalent to choosing a specific initial velocity. However, it turns out that the principle of maximizing proper time is a local concept, meaning that there may be more than one geodesic path between a pair of events. I.e. geodesics are local maxima, not necessarily global maxima. In such cases, the principle of maximizing proper time is ambiguous, but the different geodesics will each correspond to different initial velocities.

• Thanks. Now if you pick an initial point and an initial velocity (here vt, 0, 0, 0), I am not still clear why that leads to a downwards geodesic? Why not in some other direction? Jul 26, 2021 at 13:13
• @Nayeem1 it doesn't necessarily, there are many upward geodesics. In fact, the initial velocity determines if the geodesic is upwards or downwards
– Dale
Jul 26, 2021 at 13:17
• Can you please clarify how an initial velocity only in the time direction determines a downwards geodesic? Jul 26, 2021 at 13:20
– Dale
Jul 26, 2021 at 14:57
• @Nayeem1: This same issue arises in conventional Lagrangian mechanics. Hamilton's principle starts from the assumption of fixed boundary values, but we then turn around and use the resulting Euler-Lagrange equations with specified initial values instead. This bit of interpretational jiu-jitsu has been discussed on this forum here and here and here, among other places. Jul 26, 2021 at 15:28

Here is an explicit calculation which might yield some insight. If we take the flat-earth approximation (which in Newtonian mechanics corresponds to $$\vec F_g = - mg \hat z$$), then an observer standing at rest on the surface of the Earth can be modeled as a Rindler observer undergoing constant proper acceleration $$a_0\approx 9.8 \frac{\text{m}}{\text{s}^2}$$ in the $$+\hat z$$ direction. I describe the geodesic equations in my answer here; the main takeaway is that the geodesic equations in this coordinate system are

$$\frac{d^2 t}{d\tau^2} + \frac{a_0/c^2}{1+a_0 x/c^2}\left(\frac{dz}{d\tau}\right)^2 = 0$$ $$\frac{d^2x}{d\tau^2} = \frac{d^2 y}{d\tau^2} = 0$$ $$\frac{d^2 z}{d\tau^2} + a_0\frac{1+ \frac{a_0 z}{c^2}}{1+\frac{a_0z}{c^2} - \frac{v^2}{c^2}} = 0$$

"Near" the surface of the Earth ($$a_0z/c^2 \ll 1 \implies z \ll \$$5.7 trillion miles) and for non-relativistic speeds ($$v/c \ll 1$$), the geodesic equations reduce to

$$t = \tau$$ $$\frac{d^2 x}{dt^2} = \frac{d^2 y}{dt^2} = 0$$ $$\frac{d^2 z}{dt^2} + a_0 = 0$$

Given a set of initial conditions, the geodesic followed by a projectile is uniquely defined; for a projectile starting at rest (initial 4-velocity $$u = (1,0,0,0)$$), that geodesic is given by $$\mathbf x = (t,x_0,y_0, z_0 - \frac{1}{2} a_0 t^2)$$.

As a word of warning, the Rindler metric is generally understood to describe (a portion of) flat Minkowski spacetime as viewed by a uniformly accelerating observer. As such, if you were to compute the components of the Riemann curvature tensor in these coordinates, you'd find that they are all equal to zero.

As a more realistic calculation, you might consider the Newtonian limit of the Swarzschild metric. This limit gives rise to the Newtonian gravitational force $$\vec F_g = -\frac{GMm}{r^2} \hat r$$. Because this field is not uniform, the components of the Riemann tensor are not zero, and the resulting geodesics can be more honestly attributed to a real gravitational field. That being said, if you take the appropriate limits then the ultimate calculation will end up being the same.